Question

31. The sum of squares of two numbers is 80 and
the square of their difference is 36. The product of
the two numbers is
(a) 22
(b) 44
(c) 58
(d) 116​

Answer 1

Step-by-step explanation:

The answer is 22.

Let the two numbers be x, and y.

The conditions given are:

The sum of squares of two numbers is 80.

x²+y²=80

The square of difference between the two numbers is 36.

(x-y)²=36

x²-2xy+y²=36

Take the second condition, and derive a value for x².

x²-2xy+2xy+y²-y²=36+2xy-y²

x²=-y²+2xy+36

Replace x² in the first condition with the derived value.

x²+y²=80

(-y²+2xy+36)+y²=80

y²-y²+2xy+36=80

2xy+36–36=80–36

2xy÷2=44÷2

xy=22

Thus the product of the two numbers (x,y) is 22.

Answer 2

Answer:

a)22

Step-by-step explanation:

we have,2 numbers be x and y

first condition

sum of squares of 2 numbers is 80

$x {}^{2} + y {}^{2} = 80$

square difference of 2 numbers is 36

$(x - y) {}^{2} = 36 \\ x { }^{2} -2 xy + y {}^{2} { = 36}$

In Second condition, take derive a value for

$x {}^{2}$

$x { }^{2} { - 2xy} + 2xy + y {}^{2} - y {}^{2} = 36 + 2xy - y {}^{2}$

Replace

$x {}^{2}$

in first condition with derived value

$x {}^{2} + y {}^{2} = 80 \\ ( - y {}^{2} + 2xy + 36) + y {}^{2} = 80 \\ y {}^{2} - y {}^{2} + 2xy + 36 = 80 \\ 2xy + 36 - 36 = 80 - 36 \\ 2xy \div 2 = 44 \div 2 \\ xy = 22$

PRODUCT OF 2 NUMBERS=

$(x.y) = 22$